the ppt is based on the composition of forces and have types.

2. Presented By:
Amenah Gondal
Presented To:
Ma’am Mehwish
Class:
B.S.Ed (I)
Topic:
Composition of Forces

4.  Composition of forces
 Components of Forces
 Composition of Concurrent Forces
 Geometrical Methods
 Analytical Method
 (𝜆, 𝜇) Theorem
 Examples

5. The basic problem of statics consists in the reduction of a
given system of forces to the simplest system which will be
equivalent to the original system. Such a reduction is called
composition of forces.

6.  When two forces 𝑃 and 𝑄 act on a particle then their
resultant 𝑅 can be find out by using Law of Parallelogram.
The forces 𝑃 and 𝑄 are called components of the resultant.
O
𝑃
𝑄
R

7.  Forces whose lines of actions intersect at one point called
concurrent forces.
 Such forces always have a resultant through the point of
concurrence.
 This resultant can be found either geometrically or
analytically.
 The two methods are:
 Geometrical method
 Analytical method

8. It consists of three types of forces:
 Two Forces
 Three non-coplanar Forces
 System of Forces

9.  Two Forces 𝑃1 𝑎𝑛𝑑 𝑃2
 Find Resultant 𝑅 using
parallelogram law of forces
 From figure
𝑂𝐶 = 𝑂𝐴 + 𝐴𝐶
∴ 𝑅 = 𝑃1 + 𝑃2
 i.e., the resultant of two concurrent forces is given by their
vector sum
𝑅
𝑃1
𝑃2
O A
C

10.  Alternatively, Resultant 𝑅can be calculated using law of
trigonometry:
Using law of cosine in ∆𝑂𝐴𝐵
c2= 𝑎2+𝑏2 − 2𝑎𝑏cos α
R2= 𝑃1
2
+𝑃2
2
-2P1 P2cos (180°-α)
R= 𝑃1
2
+ 𝑃1
2
− 2𝑃1 𝑃2cos (180° − 𝛼)
R= 𝑃1
2
+ 𝑃1
2
+ 2𝑃1 𝑃2cos𝛼
𝑅
𝑃1
𝑃1
𝑃2
𝑃2
O A
BC
𝜶 𝜷
𝜸

11. The angles 𝛾 𝑎𝑛𝑑 𝛽 which the resultant makes with
component forces are found by the law of sine. In ∆𝑂𝐴𝐵
𝒂
𝒔𝒊𝒏𝜶
=
𝒃
𝒔𝒊𝒏𝜷
=
𝒄
𝒔𝒊𝒏𝜸
𝑷 𝟏
𝒔𝒊𝒏𝜸
=
𝑷 𝟐
𝒔𝒊𝒏𝜷
=
𝑹
𝒔𝒊𝒏(𝟏𝟖𝟎° − 𝜶)
⇒
𝑷 𝟏
𝒔𝒊𝒏𝜸
=
𝑷 𝟐
𝒔𝒊𝒏𝜷
=
𝑹
𝒔𝒊𝒏𝜶

12.  Non-coplanar Forces are 𝑃1, 𝑃2 𝑎𝑛𝑑 𝑃3.
 To find Resultant construct parallelepiped whose edges are
𝑂𝐴,𝑂𝐵 𝑎𝑛𝑑 𝑂𝐶.
 By the law of parallelogram
of Forces
𝑂𝐴 + 𝑂𝐵=𝑂𝐸
And applying the same law again
𝑂𝐸 + 𝑂𝐶= 𝑂𝐷
𝑃1
𝑃2
𝑃3
A
B
C
D
E
O
𝑅

13. Put value of 𝑂𝐸
𝑂𝐴 + 𝑂𝐵+𝑂𝐶=𝑂𝐷
⇒ 𝑃1 + 𝑃2+𝑃3= diagonal
of the parallelepiped through O.
Hence the resultant 𝑅 of three concurrent non-coplanar
forces, acting at O, is represented by the diagonal, drawn
through O, of a parallelepiped with the given forces.
We can express the resultant of construction for the resultant
by the vector equation
𝑖=1
3
𝑃𝑖 = 𝑅

14.  For more than two forces say n .
 The resultant will be obtained by repeated application of
parallelogram law of forces.
𝑖=1
𝑛
𝑃𝑖 = 𝑅
where i= 1,2,3,…………….,n
O
𝑃3
𝑃1
𝑃2
𝑃4

15. The analytical method is based on the following
theorem:
“If R is the resultant of the forces 𝑃1, 𝑃2, 𝑃3……………..𝑃𝑛
then the component of R in any direction is equal to the
algebraic sum of the components of 𝑃1 +
𝑃2+𝑃3……………..𝑃𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒
𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛.”
O y
x
z
𝑃1
𝑃2
𝑃𝑛
𝑃3
𝑃4
R

16. To prove the theorem we take a unit vector 𝑢 along the
direction in which the components are taken. Since
R= 𝑖=1
𝑛
𝑃𝑖
We multiply this equation scalarly by 𝑢
R. 𝑢=(𝑃1 + 𝑃2 + 𝑃3……………..+𝑃𝑛). 𝑢
By using distributive property of vectors
R. 𝑢=𝑃1. 𝑢 + 𝑃2. 𝑢 + 𝑃3. 𝑢+……………..+𝑃𝑛. 𝑢
=(sum of components of component vectors along 𝑢)
R . 𝑢 = ( 𝑖=1
𝑛
𝑃𝑖). 𝑢 = 𝑖=1
𝑛
𝑃𝑖 𝑢
which completes the proof.

17.  x-component of 𝑅= sum of x-components of forces 𝑃1 +
𝑃2+𝑃3……………..𝑃𝑛
Rcos𝛼 =X = 𝑖=0
𝑛
𝑃𝑖 𝑐𝑜𝑠𝛼𝑖 (i)
 y-component of 𝑅= sum of y-components of forces 𝑃1 +
𝑃2+𝑃3……………..𝑃𝑛
Rcos𝛽 =Y = 𝑖=0
𝑛
𝑃𝑖 𝑐𝑜𝑠𝛽𝑖(ii)
 z-component of 𝑅= sum of z-components of forces 𝑃1 +
𝑃2+𝑃3……………..𝑃𝑛
Rcos𝛾 =Z = 𝑖=0
𝑛
𝑃𝑖 𝑐𝑜𝑠𝛾𝑖(iii)

18. Squaring and adding eq. i. ,ii and iii
 R2cos𝛼 + 𝑅2 𝑐𝑜𝑠𝛽+ 𝑅2cos𝛾
=( 𝑖=0
𝑛
𝑃𝑖 𝑐𝑜𝑠𝛼𝑖)2+( 𝑖=0
𝑛
𝑃 𝑖 𝑐𝑜𝑠𝛽𝑖)2+( 𝑖=0
𝑛
𝑃𝑖 𝑐𝑜𝑠𝛾𝑖)2
 R2(cos𝛼 + 𝑐𝑜𝑠𝛽+ cos𝛾)
= ( 𝑖=0
𝑛
𝑃𝑖 𝑐𝑜𝑠𝛼𝑖)2+( 𝑖=0
𝑛
𝑃 𝑖 𝑐𝑜𝑠𝛽𝑖)2+( 𝑖=0
𝑛
𝑃𝑖 𝑐𝑜𝑠𝛾𝑖)2
 R2=X2+Y2+Z2
 R= 𝑋2 + 𝑌2 + 𝑍2
which determine the magnitude of the resultant R.

19. Resultant Angle of 𝛼,𝛽 and 𝛾 can be calculated by the
following results:
As Rcos𝛼 =X ⇒ 𝑐𝑜𝑠𝛼 =
𝑋
𝑅
Similarly Rcos𝛽 =Y ⇒ 𝑐𝑜𝑠𝛽 =
𝑌
𝑅
Rcos𝛾 =Z ⇒ 𝑐𝑜𝑠𝛾 =
𝑍
𝑅
which determines the direction of the resultant R.

20. Forces P,Q, R act at a point parallel to the sides of triangle
ABC taken in the same order. Show that the magnitude of the
resultant is
(𝑃2
+ 𝑄2
+ 𝑅2
− 2𝑄𝑅𝑐𝑜𝑠𝐴 − 2𝑅𝑃𝑐𝑜𝑠𝐵 − 2𝑃𝑄𝑐𝑜𝑠𝐶)1 2
Sol:
Take rectangular axes Ox, Oy with Ox parallel to the
side BC of the triangle ABC
Now
< 𝑃𝑂𝑄 = 180° − 𝐶
< 𝑃𝑂𝑅 = 180° + 𝐵
180° − 𝐶180° + 𝐵
180° − 𝐴
𝑃
A
𝑄𝑅
CB
x
o
y
𝑃
𝑅
𝑄

21. If X and Y are the components of the resultant along axes,
X=P + Q cos(180° − 𝐶) + 𝑅𝑐𝑜𝑠(180° + 𝐵)
= P−Q cos𝐶 − 𝑅𝑐𝑜𝑠𝐵---------- I
Y=Q sin(180° − 𝐶) + 𝑅𝑠𝑖𝑛(180° + 𝐵)
= Qsin𝐶 − 𝑅𝑠𝑖𝑛𝐵---------------II
Hence squaring and adding I and II
𝑋2 + 𝑌2 =
(𝑃 − 𝑄 𝑐𝑜𝑠𝐶 − 𝑅𝑐𝑜𝑠𝐵)2+(𝑄𝑠𝑖𝑛𝐶 − 𝑅𝑠𝑖𝑛𝐵)2= 𝑃2 +
𝑄2 𝑐𝑜𝑠2 𝐶 + 𝑅2 𝑐𝑜𝑠2 𝐵 − 2𝑃𝑄𝑐𝑜𝑠𝐶 + 2𝑄𝑅( 𝑐𝑜𝑠𝐵𝑐𝑜𝑠𝐶 −

22. = 𝑃2+𝑄2 + 𝑅2 + 2𝑄𝑅𝑐𝑜𝑠(180° − 𝐴) − 2𝑅𝑃𝑐𝑜𝑠𝐵 − 2𝑃𝑄𝑐𝑜𝑠𝐶
= 𝑃2+𝑄2 + 𝑅2 + 2𝑄𝑅𝑐𝑜𝑠𝐴 − 2𝑅𝑃𝑐𝑜𝑠𝐵 − 2𝑃𝑄𝑐𝑜𝑠𝐶
Taking square root on both sides
𝑋2 + 𝑌2
= 𝑃2+𝑄2 + 𝑅2 + 2𝑄𝑅𝑐𝑜𝑠𝐴 − 2𝑅𝑃𝑐𝑜𝑠𝐵 − 2𝑃𝑄𝑐𝑜𝑠𝐶
= (𝑃2
+ 𝑄2
+ 𝑅2
− 2𝑄𝑅𝑐𝑜𝑠𝐴 − 2𝑅𝑃𝑐𝑜𝑠𝐵 − 2𝑃𝑄𝑐𝑜𝑠𝐶)1 2

23. "If two concurrent forces are represented by 𝜆OA and
𝜇OB then their resultant is given by (𝜆 + 𝜇)𝑂𝐶, where C
divides AB so that AC: CB= 𝜇: 𝜆.”
Proof:
In order to prove this theorem we proceed as follows:
In the ΔOAC
𝑂𝐴=𝑂𝐶+𝐶𝐴 ------------ (I)
In the ΔOBC
𝑂𝐵=𝑂𝐶+𝐶𝐵 ------------ (II)
𝐵
𝐴
𝑂
𝐶𝝀𝑂𝐴
𝜇𝑂𝐵
𝜇
𝝀

24. Multiplying (I) by 𝝀 and (II) by 𝝁
𝜆𝑂𝐴=𝜆𝑂𝐶+𝜆𝐶𝐴 ---------------- (III)
𝜇𝑂𝐵=𝜇𝑂𝐶+𝜇𝐶𝐵 ----------------- (IV)
Now adding (III) and (IV)
𝜆𝑂𝐴 + 𝜇𝑂𝐵= (𝜆𝑂𝐶+𝜆𝐶𝐴 ) + (𝜇𝑂𝐶 + 𝜇𝐶𝐵) ----------- (V)
𝜆𝑂𝐴 + 𝜇𝑂𝐵=(𝜆 + 𝜇)𝑂𝐶+ 𝜆𝐶𝐴 + 𝜇𝐶𝐵-------------- (VI)
As , it is given that AC:CB=𝜇: 𝜆
𝐴𝐶
𝐶𝐵
=
𝜇
𝜆
⟹ 𝜆𝐴𝐶=𝜇𝐶𝐵
𝜇𝐶𝐵 − 𝜆𝐴𝐶=0
𝜇𝐶𝐵 + 𝜆𝐶𝐴=0
∴ 𝑠𝑖𝑛𝑐𝑒 − 𝜆𝐴𝐶 = 𝜆𝐶𝐴

25. Then eq (VI) becomes
𝝀𝑶𝑨 + 𝝁𝑶𝑩=(𝝀 + 𝝁)𝑶𝑪
which completes a theorem
In case
𝜆 = 𝜇 = 1
Then (𝜆, 𝜇) theorem becomes
𝑂𝐴 + 𝑂𝐵= 2𝑂𝐶
Where C is the middle point of AB .This in fact the result
given by the parallelogram of forces since OC is half of the
diagonal of the parallelogram.

26. Forces P,Q, R act at a point O and their resultant is R. If
any transversal cuts the lines of action of the forces in
the points A, B and C respectively, prove that
𝑷
𝑶𝑨
+
𝑸
𝑶𝑩
=
𝑹
𝑶𝑪
Sol:
We can use the (𝜆, 𝜇) theorem to get the required results:
As 𝑃 = 𝑃 𝑂𝐴 = 𝑃
𝑂𝐴
𝑂𝐴
=
𝑃
𝑂𝐴
𝑂𝐴
𝑄 = 𝑄 𝑂𝐵 = 𝑄
𝑂𝐵
𝑂𝐵
= (
𝑄
𝑂𝐵
)𝑂𝐵
𝑅 = 𝑅 𝑂𝐶 = 𝑅
𝑂𝐶
𝑂𝐶
= (
𝑅
𝑂𝐶
)𝑂𝐶
B
C
A
O
𝑄
𝑃
𝑅
L

27. Using the (𝜆, 𝜇) theorem, we have
𝑃 + 𝑄 =
𝑃
𝑂𝐴
𝑂𝐴 +
𝑄
𝑂𝐵
𝑂𝐵
=(
𝑃
𝑂𝐴
+
𝑄
𝑂𝐵
) 𝑂𝐴 + 𝑂𝐵
= (
𝑃
𝑂𝐴
+
𝑄
𝑂𝐵
) 𝑂𝐶
∴ (𝑂𝐴 + 𝑂𝐵 = 𝑂𝐶)
Where C is a point lie on AB such that
AC: CB =
𝑄
𝑂𝐵
:
𝑃
𝑂𝐴
As we know that
𝑃 + 𝑄 = 𝑅
Then
𝑃
𝑂𝐴
+
𝑄
𝑂𝐵
𝑂𝐶 = (
𝑅
𝑂𝐶
)𝑂𝐶

28. which gives
𝑃
𝑂𝐴
+
𝑄
𝑂𝐵
= (
𝑅
𝑂𝐶
)
Hence proved
𝑃
𝑂𝐴
+
𝑄
𝑂𝐵
=
𝑅
𝑂𝐶